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Solve for $x.$$$\log (x+3)+\log (x+2)=\log 20$$

$$2$$

Algebra

Chapter 4

Exponential and Logarithmic Functions

Section 6

Properties of Logarithmic Functions

Missouri State University

Campbell University

Harvey Mudd College

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Ah, I like to write out the implied base. You don't have to in this problem, but the implied bases. 10. When there's no base written, you know we have a plus log of X plus two. But again, I like to write the base and then it equals log base 10 of 20. So the main idea in this problems this plus sign we can combine the two logs together using multiplication. Now I'm going to go ahead and distribute X to each term. So we're looking at X squared plus two X and then distribute three in there, plus three actual two x plus three xs five x and then three times to it's six. Some people use the phrase foil. Ah, I know some math teachers don't like foiling eso now. What we can do is because we have logged based, tenable sides. They actually cancel each other out. And what I'm really doing is taking 10 thio each power on either side. And now I'm ready to solve this because it's a quadratic and we solve quadratic by getting this equal to zero. Um, so it's just X squared. Plus five x plus six equals 20 so I can just subtract 20 over. I'm looking at X squared plus five x minus 14. Well, factors of 14 that add to be five. Sorry, negative 14 would be negative two and positive seven. So we're looking at X minus two X plus seven. So are two answers in the problem. If I use a zero product property and solve is the X equals positive two or exit goes negative. Seven. Except we cannot log negatives. So if I plug in to and for here, two plus three is five. That's good. Weaken long a positive Now two plus two is four. That's good. We can log a positive, but if we look at negative seven negative seven plus three is negative. Four right here. Can't do that. So this answer is not It's extraneous. It does not work. Exit goes to is your only answer.

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